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The divisor function $\tau (n)={\sum }_{d\mid n}1$ counts the positive integer divisors of a positive integer $n$. In particular $\tau (1)=1$, because $1$ has just one positive integer divisor. More generally the notation ${\sigma }_k(n)={\sum }_{d\mid n}{d}^k$ gives $\tau ={\sigma }_0$ and the sum-of-divisors function ${\sigma }_1$. Thus the divisor function counts the positive integer divisors, while ${\sigma }_1$ adds their sizes. ^[ref]

If the prime factorization is $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, thenThe short proof is a count: each positive integer divisor has a unique form ${\prod }_{k=1}^r{p}_k^{b_k}$, where $0\le b_k\le a_k$. There are $a_k+1$ choices at each prime number, and choosing them gives every positive integer divisor exactly once. For $n=1$ there are no prime factors and the product is $1$.

For $12=2^2\cdot 3$, choose $i\in \{0,1,2\}$ and $j\in \{0,1\}$ to obtain $2^i{3}^j$. The six values are $1,2,4,3,6,12$; sorted, the positive integer divisors are $1,2,3,4,6,12$. Therefore $\tau (12)=(2+1)(1+1)=6$. The diagram makes the independent exponent choices visible.

This count is useful because a known prime factorization gives $\tau (n)$ without testing every smaller positive integer for divisibility. It also proves the divisor function is a multiplicative function: coprime integers have disjoint prime factors, so their exponent choices combine. For example $\tau (6)=4=\tau (2)\tau (3)$, but $\tau (4)=3\ne 4=\tau (2{)}^2$, so it is not a completely multiplicative function. Equivalently, $\tau =1\ast 1$ under Dirichlet convolution, counting ordered pairs of positive integers whose product is $n$.

The divisor sum of an arithmetic function $f$ assigns to each positive integer $n$ the sumOnly positive integer divisors are included, so $F(1)=f(1)$. The weights $f(d)$ can count the integer divisors, measure their sizes, or record other arithmetic information: the constant-one arithmetic function gives the divisor function, whereas the identity arithmetic function gives the sum-of-divisors function.

Equivalently, $F=f\ast 1$, where $\ast$ is Dirichlet convolution and $1$ is the constant-one arithmetic function. If $f$ is a multiplicative function, then $F$ is a multiplicative function by Multiplicativity of Dirichlet convolution. ^[ref]

Here is the mechanism. For positive coprime integers $m,n$, every positive integer divisor of $mn$ has a unique form $ab$ with $a\mid m$ and $b\mid n$; moreover, $a,b$ are coprime integers. Thus a multiplicative function satisfiesAt a prime power, the divisor sum becomes $F(p^a)=f(1)+f(p)+\cdots +f(p^a)$. This turns a calculation over many integer divisors into short local calculations at prime powers.

The Möbius function assigns a cancellation weight to each positive integer:Thus a squarefree integer contributes a sign determined by its number of distinct prime factors; a repeated prime factor makes the weight zero. This packages the alternating signs of inclusion-exclusion into an arithmetic function. ^[ref]

For example, $\mu (2)=\mu (3)=-1$, $\mu (6)=1$, and $\mu (30)=(-1{)}^3=-1$, since $30=2\cdot 3\cdot 5$. In contrast, $\mu (4)=\mu (12)=\mu (60)=0$: all three have the square of the prime number $2$ as a divisor. On prime powers, $\mu (p)=-1$ and $\mu (p^a)=0$ for $a\ge 2$.

The Möbius function is a multiplicative function: for coprime integers, their prime factors are disjoint, so either a squared prime factor forces both sides of $\mu (mn)=\mu (m)\mu (n)$ to vanish, or the counts of distinct prime factors add and their signs multiply. It is not a completely multiplicative function, since $\mu (4)=0\ne 1=\mu (2{)}^2$. Its Möbius divisor-sum identity makes it the inverse of the constant-one arithmetic function under Dirichlet convolution: $\mu \ast 1=\epsilon$, the Identity for Dirichlet convolution. This is why the small set of weights $-1,0,1$ can undo an entire divisor sum.

An arithmetic function $f$ is a multiplicative function when $f(1)=1$ and $f(mn)=f(m)f(n)$ for all coprime integers $m,n>0$, meaning their greatest common divisor is $1$. The condition $f(1)=1$ excludes the identically zero function. This definition makes no assertion about a pair whose greatest common divisor exceeds $1$. ^[ref]

For the divisor function, the integer divisors of $2$ are $1,2$, those of $3$ are $1,3$, and those of $6$ are $1,2,3,6$. HenceHere $2$ and $3$ are coprime integers. In contrast, $\tau (4)=3\ne 4=\tau (2{)}^2$, because the integer divisors of $4$ are $1,2,4$. Thus the divisor function is a multiplicative function but not a completely multiplicative function.

The practical gain is that prime factorization reduces all values to values at prime powers. For distinct prime numbers $p_1,\dots ,p_r$ and positive exponents $a_1,\dots ,a_r$,For instance, $f(12)=f(4)f(3)$ since $4$ and $3$ are coprime integers; replacing $f(4)$ by $f(2{)}^2$ additionally requires a completely multiplicative function. The short proof repeatedly applies the defining rule to prime powers belonging to different prime numbers, which are coprime integers.

For an arithmetic function, multiplicativity means $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m,n$ are positive coprime integers. It allows values of an arithmetic function to be assembled from its values on prime powers, as explained under multiplicative function.

The Dirichlet convolution of two multiplicative functions is a multiplicative function. For positive coprime integers $m,n$, every positive integer divisor of $mn$ has a unique form $de$ with $d\mid m$ and $e\mid n$, by prime factorization. The pairs $d,e$ and $m/d,n/e$ are each coprime integers, so the defining rules for the two multiplicative functions separate the sum:Also $(f\ast g)(1)=f(1)g(1)=1$. This short argument proves multiplicativity of both the divisor function $1\ast 1$ and the sum-of-divisors function $\mathrm{id}\ast 1$. The result need not be a completely multiplicative function, even when both inputs are: $1\ast 1=\tau$ fails at $2\cdot 2$. ^[ref]

The radical of an integer $n>0$ is the product of its distinct prime factors, each included exactly once. If the prime factorization is $n=p_1^{a_1}\cdots p_r^{a_r}$ with distinct prime numbers $p_i$ and positive integer exponents $a_i$, thenThus the radical of an integer remembers which prime factors occur and forgets their prime valuations. At $n=1$, the empty product gives $\mathrm{rad}(1)=1$. The domain of a function here is the positive integers; zero is excluded. ^[ref]

For example, $72=2^3\cdot 3^2$ has distinct prime factors $2,3$, so $\mathrm{rad}(72)=2\cdot 3=6$. Similarly,using the prime factorizations $30=2\cdot 3\cdot 5$, $32=2^5$ and $36=2^2\cdot 3^2$. The prime omega function counts the retained prime factors, while the total number of prime factors counts their occurrences before repetitions are removed.

The radical of an integer is its largest squarefree divisor. Indeed, a squarefree divisor chooses at most one copy of each prime factor, so it divides $\mathrm{rad}(n)$; the radical of an integer itself is a squarefree divisor. Consequently,The last statement follows because every nonzero prime valuation is then one. It includes $n=1$, a squarefree integer. Applying the radical of an integer twice also changes nothing: $\mathrm{rad}(\mathrm{rad}(n))=\mathrm{rad}(n)$.

For any positive integers $a,b$, the exact product rule isTo prove it, a prime factor dividing just one input occurs once in the numerator and not in the denominator. A prime factor dividing both occurs twice in the numerator and once in the denominator, since it divides their greatest common divisor. Each prime factor of $ab$ therefore remains exactly once. For coprime integers, the denominator is $\mathrm{rad}(1)=1$, proving the radical of an integer is a multiplicative function. It is not a completely multiplicative function: $\mathrm{rad}(4)=2\ne 4=\mathrm{rad}(2{)}^2$.

For a worked example with shared prime factors, take $a=12$ and $b=18$. Their greatest common divisor is $6$, and $ab=216=2^3\cdot 3^3$, soThis explains why prime support of an integer is useful: questions about the presence of prime factors can discard their multiplicities. For example, the squarefree divisors of $72$ are exactly the integer divisors of $\mathrm{rad}(72)=6$, namely $1,2,3,6$. Also $\gcd (m,n)=1$ if and only if $\gcd (m,\mathrm{rad}(n))=1$, because coprimality depends only on shared prime factors.

A further characterization uses mutual divisibility of powers: for positive integers $a,b$,If their prime supports of an integer agree, sufficiently large $k,\ell$ make all the required prime valuations large enough. Conversely, these two divisibility relations force each prime factor of either input to divide the other. For $72$ and $36$, one can take $k=2$ and $\ell =1$: $72\mid 36^2$ and $36\mid 72$.

The sum-of-divisors function adds all positive integer divisors of a positive integer, including the number itself:It is the divisor sum of the identity arithmetic function $\mathrm{id}(n)=n$, so $\sigma =\mathrm{id}\ast 1$ under Dirichlet convolution. Both factors are multiplicative functions, hence so is the sum-of-divisors function. The notation ${\sigma }_1$ distinguishes it from other weighted divisor sums. ^[ref]

For a prime number $p$ and a nonnegative exponent $a$, the positive integer divisors of the prime power $p^a$ are precisely $1,p,\dots ,p^a$. Their finite geometric series givesFor example, $\sigma (8)=1+2+4+8=15=(2^4-1)/(2-1)$. For a prime factorization $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, multiplicativity givesFor $n=1$, this is an empty product, equal to $1$. ^[ref]

The distributive property makes the formula especially beautiful: choosing one term from each factor produces each positive integer divisor exactly once, by the fundamental theorem of arithmetic. For $12=4\cdot 3=2^2\cdot 3$, the choices form a three-by-two grid with entries $2^i{3}^j$, where $0\le i\le 2$ and $0\le j\le 1$. Adding its rows givesThe divisor function counts six cells; the sum-of-divisors function adds their six values. Once the prime factorization is known, the product formula avoids testing every candidate integer divisor. Shared prime factors require care: $\sigma (4)=7\ne 9=\sigma (2{)}^2$, so the sum-of-divisors function is not a completely multiplicative function.

Subtracting the number itself gives the proper-divisor sum $s(n)=\sigma (n)-n$. Comparing $s(n)$ with $n$ distinguishes perfect numbers, deficient numbers, and abundant numbers. For $6$, $s(6)=1+2+3=6$; for $28$, $s(28)=1+2+4+7+14=28$. Both are perfect numbers. The calculation $s(8)=1+2+4=7<8$ makes $8$ a deficient number, while $s(12)=1+2+3+4+6=16>12$ makes $12$ an abundant number. These comparisons turn the sizes of integer divisors into a classification of positive integers. ^[ref]

This connection is ancient: Euclid's Elements, Book IX, Proposition 36, constructs a perfect number by multiplying the last term of $1,2,\dots ,2^{k-1}$ by their sum, when that sum is a prime number. In modern notation, put $q=2^k-1$ and $N=2^{k-1}q$, with $k\ge 2$ and $q$ a prime number. The sum-of-divisors function proves the construction immediately:Consequently the proper-divisor sum is $s(N)=N$. Taking $k=2,3$ produces the perfect numbers $6,28$. ^[ref]